Let $G \cong \mathbb{Z}_{360} \times \mathbb{Z}_{120} \times \mathbb{Z}_{75} \times \mathbb{Z}_{3}$.
I want to find the number of subgroups $H \le G$ of order $25$.
First I found out that there are $124$ elements of order $5$ and $500$ elements of order $25$ in $G$.
I want to divide the subgroups to ones that are isomorphic to $\mathbb{Z}_{5} \times \mathbb{Z}_{5}$ and those that are isomorphic to $\mathbb{Z}_{25}$.
I start with $\mathbb{Z}_{25}$, so any subgroup of $G$ that is isomorphic ti it is generated by an element of order $25$. Now here is where I am not sure what to do. I saw someone calculate the amount by doing $\frac{500}{\varphi(25)}$ but I can't understand how did he get to it.
The same goes about groups of order $5$.
Help would be appreciated